Lantern Melody Chain Length

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Every evening the lantern orchestra prepares a list of melody brightness ratings arranged in the exact order they will attempt to rehearse To keep the lighting transitions smooth the conductor picks a subsequence of songs whose brightness never dips each chosen piece must be at least as bright as the one before it The conductor wants to know the maximum length of such a non-decreasing chain that can be performed in order without rearranging the original playlist The result helps decide how many lantern controllers need to staff the risers and which musicians get rehearsal priority Your routine receives an array of integers representing brightness ratings You may skip songs but cannot change their order and duplicates are welcome because equal brightness keeps the ambience balanced A single song always forms a valid chain of length one and an empty playlist should yield zero Use dynamic programming to determine the longest non-decreasing subsequence length without modifying the original array This efficient tally gives the conductor confidence that the show can glide from soft glow to radiant finale without jolting the audience Because this is a counting optimization-style challenge dynamic programming is usually the safest approach define what each index or state means initialize valid base states and make transitions explicit If the problem uses modulo arithmetic apply modulo at every accumulation step so large intermediate totals never corrupt the final answer At hard difficulty hidden tests usually combine scale with

Examples

Example 1

Input

melodies = [5,4,3]

Output

Explanation

Only individual songs preserve non-decreasing order.

Example 2

Input

melodies = []

Output

Explanation

With no melodies, the chain length is zero.

Example 3

Input

melodies = [1,3,2,3,4]

Output

Explanation

One optimal chain is [1,2,3,4].

Algorithm Flow

Recommendation Algorithm Flow for Lantern Melody Chain Length - Budibadu

Best Answers

java

import java.util.*;
class Solution {
    public int calculate_chain_length(int[] notes) {
        if (notes.length == 0) return 0;
        int[] tails = new int[notes.length];
        int len = 0;
        for (int x : notes) {
            int i = 0, j = len;
            while (i != j) {
                int m = (i + j) / 2;
                if (tails[m] < x) i = m + 1;
                else j = m;
            }
            tails[i] = x;
            if (i == len) len++;
        }
        return len;
    }
}

javascript

function calculate_chain_length(notes) {
    if (notes.length === 0) return 0;
    const tails = [];
    for (let x of notes) {
        let low = 0, high = tails.length - 1;
        let idx = tails.length;
        while (low <= high) {
            let mid = Math.floor((low + high) / 2);
            if (tails[mid] >= x) {
                idx = mid;
                high = mid - 1;
            } else low = mid + 1;
        }
        if (idx < tails.length) tails[idx] = x;
        else tails.push(x);
    }
    return tails.length;
}

php

function calculate_chain_length($notes) {
    if (empty($notes)) return 0;
    $tails = [];
    foreach ($notes as $x) {
        $low = 0; $high = count($tails) - 1;
        $idx = count($tails);
        while ($low <= $high) {
            $mid = floor(($low + $high) / 2);
            if ($tails[$mid] >= $x) {
                $idx = $mid;
                $high = $mid - 1;
            } else $low = $mid + 1;
        }
        if ($idx < count($tails)) $tails[$idx] = $x;
        else $tails[] = $x;
    }
    return count($tails);
}

python

import bisect
def calculate_chain_length(notes):
    if not notes: return 0
    tails = []
    for x in notes:
        idx = bisect.bisect_left(tails, x)
        if idx < len(tails): tails[idx] = x
        else: tails.append(x)
    return len(tails)

rust

pub fn calculate_chain_length(notes: Vec<i32>) -> i32 {
    if notes.is_empty() { return 0; }
    let mut tails = Vec::new();
    for &x in &notes {
        let idx = tails.binary_search(&x);
        match idx {
            Ok(i) => tails[i] = x,
            Err(i) => {
                if i < tails.len() { tails[i] = x; }
                else { tails.push(x); }
            }
        }
    }
    tails.len() as i32
}

typescript

function calculate_chain_length(notes: number[]): number {
    if (notes.length === 0) return 0;
    const tails: number[] = [];
    for (let x of notes) {
        let low = 0, high = tails.length - 1;
        let idx = tails.length;
        while (low <= high) {
            let mid = Math.floor((low + high) / 2);
            if (tails[mid] >= x) {
                idx = mid;
                high = mid - 1;
            } else low = mid + 1;
        }
        if (idx < tails.length) tails[idx] = x;
        else tails.push(x);
    }
    return tails.length;
}

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Lantern Melody Chain Length

Examples

Algorithm Flow

Best Answers

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